every prime ideal is radical

Let $\\mathcal{R}$ be a commutative ring and let $\\mathfrak{P}$ bea prime ideal of $\\mathcal{R}$ .

Proposition 1.

Every prime ideal $\\mathfrak{P}$ of $\\mathcal{R}$ is a radicalideal, i.e.

\\mathfrak{P}=\\operatorname{Rad(\\mathfrak{P})}

Proof.

Recall that $\\mathfrak{P}\\subsetneq\\mathcal{R}$ is a prime idealif and only if for any $a,b\\in\\mathcal{R}$

a\\cdot b\\in\\mathfrak{P}\\Rightarrow a\\in\\mathfrak{P}\\text{ or }b\\in\\mathfrak{P}

Also, recall that

\\operatorname{Rad}(\\mathfrak{P})=\\{r\\in\\mathcal{R}\\mid\\exists n\\in\\mathbb{N}%\\text{ such that }r^{n}\\in\\mathfrak{P}\\}

Obviously, we have $\\mathfrak{P}\\subseteq\\operatorname{Rad}(\\mathfrak{P})$ (just take $n=1$ ), so it remainsto show the reverse inclusion.

Suppose $r\\in\\operatorname{Rad}(\\mathfrak{P})$ , so there existssome $n\\in\\mathbb{N}$ such that $r^{n}\\in\\mathfrak{P}$ . We want toprove that $r$ must be an element of the prime ideal $\\mathfrak{P}$ . For this, we use induction on $n$ to prove thefollowing proposition:

For all $n\\in\\mathbb{N}$ , for all $r\\in\\mathcal{R}$ , $r^{n}\\in\\mathfrak{P}\\Rightarrow r\\in\\mathfrak{P}$ .

Case $n=1$ : This is clear, $r\\in\\mathfrak{P}\\Rightarrow r\\in\\mathfrak{P}$ .

Case $n$ $\\Rightarrow$ Case $n+1$ : Suppose we have provedthe proposition for the case $n$ , so our induction hypothesis is

\\forall r\\in\\mathcal{R},\\quad r^{n}\\in\\mathfrak{P}\\Rightarrow r\\in\\mathfrak{P}

and suppose $r^{n+1}\\in\\mathfrak{P}$ . Then

r\\cdot r^{n}=r^{n+1}\\in\\mathfrak{P}

and since $\\mathfrak{P}$ is a prime ideal we have

r\\in\\mathfrak{P}\\text{ or }r^{n}\\in\\mathfrak{P}

Thus we conclude, either directly or using the inductionhypothesis, that $r\\in\\mathfrak{P}$ as desired.

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